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anonymous

  • one year ago

OMG!!!!!!!!!!! HELP PLZZZZZZZZ

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @Sepeario

  3. anonymous
    • one year ago
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    @sweetburger

  4. anonymous
    • one year ago
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    @nincompoop

  5. anonymous
    • one year ago
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    @LunyMoony

  6. anonymous
    • one year ago
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    @imqwerty

  7. anonymous
    • one year ago
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    plzzzzz help

  8. anonymous
    • one year ago
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    hello??????????

  9. Jack1
    • one year ago
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    for this... u srsly need to flip a coin 50 times and fill in the table from ur results @bubbleslove1234 ... sorry, till u do we cant help u hey

  10. anonymous
    • one year ago
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    i have already done this

  11. LunyMoony
    • one year ago
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    yeah, I don't understand how they want the statistics either for the 2, or 2 each, of cats and dogs on the chart

  12. Jack1
    • one year ago
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    so what were ur answers for: HH - HT - TH - and TT? -

  13. anonymous
    • one year ago
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    hold on

  14. anonymous
    • one year ago
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    hh=11 ht=17 th=8 tt=14

  15. Jack1
    • one year ago
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    ok... so from ur screenshot: which question do u want to go through first?

  16. anonymous
    • one year ago
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    the first one and work from there

  17. Jack1
    • one year ago
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    k_ so cats are Heads, and Dogs are Tails according to the start of ur Q: having cat or dog is 50:50 "theoretical" probability so: Q1 - theoretical Probability of 2 cats OR 2 dogs = P(HH) OR P(TT) = P(HH) + P(TT) Theoretical probability of HH =0.25 Theoretical probability of TH =0.25 Theoretical probability of HT =0.25 Theoretical probability of TT =0.25 so solve = P(HH) + P(TT) @bubbleslove1234 ?

  18. anonymous
    • one year ago
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    do you know what i should get cause i am soooooooooooo confused

  19. anonymous
    • one year ago
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    oww my brain hurts

  20. Jack1
    • one year ago
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    Theoretical probability is: P(HH) =0.25 (or 25%) P(TH) =0.25 (or 25%) P(HT) =0.25 (or 25%) P(TT) =0.25 (or 25%) so Q1. solve = P(HH) + P(TT)

  21. anonymous
    • one year ago
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    okay so 25% times 50

  22. Jack1
    • one year ago
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    no, sorry u can either use percentages or use decimals, but u can't use both at the same time, sorry P(HH) = 0.25 P(TT) = 0.25 so = P(HH) + P(TT) =?

  23. anonymous
    • one year ago
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    okay

  24. Jack1
    • one year ago
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    P(HH) = 0.25 P(TT) = 0.25 so P(HH) + P(TT) = ????? @bubbleslove1234 ...?

  25. anonymous
    • one year ago
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    p(11)=0.25 p(14)=0.25

  26. arindameducationusc
    • one year ago
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    did you get the Question 1 answer? dog and cat?

  27. anonymous
    • one year ago
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    nope not yet but i need fast help cause it is due today

  28. Jack1
    • one year ago
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    no, for question 1, we only care about Theoretical results... not actual one we get from flipping a coin so Theoretical probability of 2 heads is 25% of the time... so P(HH) = 0.25 and the Theoretical probability of 2 tails is 25% of the time... so P(TT) = 0.25 so whats .25 + .25 ?

  29. anonymous
    • one year ago
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    50

  30. Jack1
    • one year ago
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    perfect. PHH + PTT = 0.50 .... or 50 % so Q1 = 50% chance

  31. anonymous
    • one year ago
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    oh so is that the answer

  32. Jack1
    • one year ago
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    Q1. chance family has 2 cats or 2 dogs? = P(CC) or P(DD) = 0.5 .... or 50% yep, answer

  33. anonymous
    • one year ago
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    for the first one

  34. Jack1
    • one year ago
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    Q2. how does 2 coins simulate which 2 pets a family has? answer: 50% chance of choosing a cat or a dog, and they chose twice (50% either cat or dog, then 50% chance either cat or dog for the 2nd pet also) so exactly as flipping a coin... twice (50% either heads or tail, then 50% chance either head or tail on the 2nd toss also)

  35. Jack1
    • one year ago
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    Q3.check if this is correct: ur results were: |dw:1439454960600:dw|

  36. Jack1
    • one year ago
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    |dw:1439455108667:dw|

  37. Jack1
    • one year ago
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    Q4. Experimental Probability In ur test, how many times did u flip the coin? @bubbleslove1234 ?

  38. anonymous
    • one year ago
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    50

  39. Jack1
    • one year ago
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    perfect now, calculate your results for P(HH) and P(TT) P(HH) = number of results that were HH / total number of flips =11/50 = what? (in decimals) P(TT) = number of results that were *TT / total number of flips =14/50 = what? (in decimals) @bubbleslove1234 ?

  40. Jack1
    • one year ago
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    sorry, typo

  41. anonymous
    • one year ago
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    P(HH) = 0.22 P(TT) =0.28

  42. Jack1
    • one year ago
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    absolutely right so: Experimental Probability of 2 cats or 2 dogs = same probability as flipping a coin 50 times and getting 2 heads or 2 tails so = P(HH) + P(TT) = 0.22 + 0.28 = ???

  43. anonymous
    • one year ago
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    0.5

  44. anonymous
    • one year ago
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    @Jack1

  45. arindameducationusc
    • one year ago
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    @bubbleslove1234 should I try?

  46. anonymous
    • one year ago
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    sure

  47. arindameducationusc
    • one year ago
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    okay for question 1, Dog/Cat 1 1 2 0 0 2 now, in the question its given 2 cats or 2 dogs, so trials are 3, so, its 2/3

  48. Jack1
    • one year ago
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    sorry my page froze... and is being glitchy can u finish @arindameducationusc please?

  49. arindameducationusc
    • one year ago
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    okay @Jack1 let me try.........

  50. arindameducationusc
    • one year ago
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    For question 2, just assign head as dog and tail as cat

  51. anonymous
    • one year ago
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    kk

  52. arindameducationusc
    • one year ago
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    now, hh=11 ht=17 th=8 tt=14 if this your data..... then count head head+tail+tail (according to question4 2 dogs and 2 cats) so, 11+14=25 now for probability divide by the frequency, so answer is 25/50

  53. anonymous
    • one year ago
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    okay

  54. anonymous
    • one year ago
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    got it

  55. anonymous
    • one year ago
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    can we move to Q5

  56. arindameducationusc
    • one year ago
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    yes sure,

  57. anonymous
    • one year ago
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    thanks

  58. arindameducationusc
    • one year ago
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    for question 5, C=Cat, D=Dog CCD CDD DDD DCC DDC CCC CDC DCD now, count where CCC and DDD is there which is 2. 2/8 , where 8 is the frequesncy. 2/8 is the probability

  59. Jack1
    • one year ago
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    hey, back and working, but keep going, happy to watch Q5: perfect, but 1 question: how did u get 2/8 for Q5... but only 2/3 for Q1? should it not be 2/4...?

  60. arindameducationusc
    • one year ago
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    Q5 and Q1 has no relation (very important)

  61. arindameducationusc
    • one year ago
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    one sec....

  62. Jack1
    • one year ago
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    no. categorically: no. Q1 and Q5 are totally related ;)

  63. arindameducationusc
    • one year ago
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    Q5 is correct.....

  64. arindameducationusc
    • one year ago
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    but Q1 I missed something.......

  65. anonymous
    • one year ago
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    what

  66. arindameducationusc
    • one year ago
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    if D=Dog C=Cat CC DD CD DC so yes it is 2/4. Thank you @Jack1

  67. Jack1
    • one year ago
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    np dude ;)

  68. arindameducationusc
    • one year ago
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    Now question 6

  69. anonymous
    • one year ago
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    kk

  70. arindameducationusc
    • one year ago
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    H=Head, T=Tail In the table given in Question, replace that with TTH THH HHH HTT HHT TTT THT HTH

  71. anonymous
    • one year ago
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    ???????

  72. anonymous
    • one year ago
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    what

  73. Jack1
    • one year ago
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    he means : flip a coin 3 times, and record the results current simulation is; flip a coin 2 times

  74. anonymous
    • one year ago
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    so what is the answer

  75. Jack1
    • one year ago
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  76. arindameducationusc
    • one year ago
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    @Jack1 Exactly ! Flipping coins 3 times.... and record observations

  77. anonymous
    • one year ago
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    only 3 times

  78. Jack1
    • one year ago
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    flip a coin... 3 times.... record the results... repeat 50 times or flip 3 coins at the same time... 50 times

  79. anonymous
    • one year ago
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    okay

  80. anonymous
    • one year ago
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    thanks

  81. arindameducationusc
    • one year ago
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    Good job @bubbleslove1234 ... Was my attempt useful @bubbleslove1234 ?

  82. anonymous
    • one year ago
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    yep

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